$\lim\limits_{n\rightarrow\infty}f(2^{n}x)=0$ implies$\lim\limits_{x\rightarrow\infty}f(x)=0?$

Question

If $f: (0,\infty)\rightarrow\mathbb{R}^{1}$ is continuous, and $\lim\limits_{n\rightarrow\infty}f(2^{n}x)=0.$ for all $x>0$. I want to prove $\lim\limits_{x\rightarrow\infty}f(x)=0$, but I have no idea.

This looks similar to the problem A classical problem about limit of continuous function at infinity and its connection with Baire Category Theorem. But when I replace $n$ with $2^{n}$, I cannot prove it. The problem is that there is no $c>0$, such that $$(c,\infty)\subset\bigcup_{n=N}^{\infty}(2^{n}a,2^{n}b).$$

I have also doubted the validity of this conclusion, and I have tried to find a counterexample, but I have not found one.I want to know if this conclusion is true, and if not, is there a counterexample? Thank you.

Answer 1

Define $g(x) = f(\exp(x))$, thus $f(x) = g(\ln x)$ and $f(2^n x) = g(n + \ln x)$. $g$ is continuous, condition become $\forall x: \lim\limits_{n\to\infty} g(n + x) = 0$, and for counterexample we need $\neg\lim\limits_{x \to \infty} g(x)= 0$.

Now, it's easy to see how to build a counterexample: let $w(x)$ be bump function: $w(0) = 1$, $w(x) = 0$ if $|x| \geq 1$, $w$ is continuous. And define $g(x) = \sum_n w((x - n + 1/n) \cdot n^3)$ - narrow bumps around $n - 1/n$, that do not intersect when shifted by integer.

It's clear that $\lim\limits_{n \to \infty} g(x)$ doesn't exist. However, every $x$ can get into at most one bump when shifted, so $g$ satisfies our condition.

Answer 2

Let $g_n(x)=x^n(1-x)n$. Note that for positive integer $n$, $g(n,0)=g(n,1)=0$ and it is continuous on $[0,1]$.

Let $f(x)$ be $0$ for $x<2$, and for positive integer $n$, for $x\in [2^n,2^{n+1}]$, let $f(x)=g_n(\log_2{(x/2^n)})$.

Since $g_n(x)<x^n * n$, for fixed $x<1$, this goes to 0 as $n$ gets big. At $1$, it’s already $0$.

As $n$ gets big $g_n(1-1/n)$ goes to $1/e$, so $f$ does not approach $0$.